suss
Xlogylogz.ylogzlogx.zlogxlogy=1
how to prove this??



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suss
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it was supposed to x to the power logylogz y to the power logzlogx z to the power logxlogy

hartnn
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\(\huge x^{\log y\log z}y^{\log z\log x}z^{\log x\log y}\)
like this ?

suss
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ah thank god..yes like that

hartnn
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consider
\(\huge \log[x^{\log y\log z}y^{\log z\log x}z^{\log x\log y}]\)
can u simplify that ?

hartnn
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use, \(\log AB=\log A+\log B \\ \log A^{B}=B\log A\)

hartnn
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you should get that =0 = ln 1

suss
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i dont understand y u multiplied the whole by log????

hartnn
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to be able to simplify

hartnn
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else, nothing can't be really done to that expression

suss
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i have done that problem in this way
\[x ^{\log y/z}y ^{\log z/x}z ^{\log x/y}\]

suss
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and i cudnt figure out the next step..... :(

hartnn
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i know, i tried that also....since i could not proceed, i thought of taking log....

suss
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if u dont mind,can u show me the simplification process...??? :)

hartnn
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\(\huge \log[x^{\log y\log z}y^{\log z\log x}z^{\log x\log y}] \\\large =\log[x^{\log y\log z}]+\log[y^{\log z\log x}]+\log[z^{\log x\log y}]\)
got this ?

hartnn
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now use
\( \\ \log A^{B}=B\log A\)

hartnn
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@suss

suss
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oh sorry man i jus lost connection
leeme check it out

suss
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y did u keep the + sign ???

suss
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@hartnn

hartnn
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because
\(\log AB=\log A+\log B \\ \)

suss
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ah finally got it .....thx a ton man

hartnn
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sure ?
u getting that = 0 ?
then write 0=log 1

suss
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yes 100%..i hav to use formula of log ab bak ther ......i really got it man

hartnn
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good :)
welcome ^_^